Let [ a_n _n = 0^ ] be a sequence such that a_0 = a_1 = 0 and a_n + 2 = 2a_n + 1 - a_n + 1…

Question

Let $$\{ {a_n}\} _{n = 0}^\infty $$ be a sequence such that ${a_0} = {a_1} = 0$ and ${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$ for all n $$\ge$$ 0. Then, $$\sum\limits_{n = 2}^\infty  {{{{a_n}} \over {{7^n}}}} $$ is equal to:

Accepted Answer

${a_{n + 2}} = 2{a_{n + 1}} - {a_n} + 1$ \& ${a_0} = {a_1} = 0$

${a_2} = 2{a_1} - {a_0} + 1 = 1$

${a_3} = 2{a_2} - {a_1} + 1 = 3$

${a_4} = 2{a_3} - {a_2} + 1 = 6$

${a_5} = 2{a_4} - {a_3} + 1 = 10$

$$\sum\limits_{n = 2}^\infty  {{{{a_n}} \over {{7^n}}} = {{{a_2}} \over {{7^2}}} + {{{a_3}} \over {{7^3}}} + {{{a_4}} \over {{7^4}}} + \,\,...} $$

$$s = {1 \over {{7^2}}} + {3 \over {{7^3}}} + {6 \over {{7^4}}} + {{10} \over {{7^5}}} + \,\,...$$

$${{{1 \over 7}s = {1 \over {{7^3}}} + {3 \over {{7^4}}} + {6 \over {{7^5}}} + \,\,...} \over {{{6s} \over 7} = {1 \over {{7^2}}} + {2 \over {{7^3}}} + {3 \over {{7^4}}} + \,\,...}}$$

$${{{{6s} \over {49}} = \,\,\,\,\,\,\,\,\,{1 \over {{7^3}}} + {2 \over {{7^4}}} + \,\,...} \over {{{36s} \over {49}} = {1 \over {{7^2}}} + {1 \over {{7^3}}} + {1 \over {{7^4}}} + \,\,...}}$$

$${{36s} \over {49}} = {{{1 \over {{7^2}}}} \over {1 - {1 \over 7}}}$$

$${{36s} \over {49}} = {7 \over {49 \times 6}}$$

$$s = {7 \over {216}}$$

Solutions · 1